Series and Areas

Let $|x| < 1$ .

$g(x) = \sum_{n = 0}^{\infty} (a_{1} + a_{2}n + a_{3}n^2)x^n = a_{1}\sum_{n = 0}^{\infty} x^n + a_{2}\sum_{n = 0}^{\infty} n x^n + a_{3}\sum_{n = 0}^{\infty} n^2 x^n$ .

$$

$a_{1}\sum_{n = 0}^{\infty} x^n + a_{2}\sum_{n = 0}^{\infty} n x^n = \dfrac{a_{1}}{1 - x} + a_{2}\sum_{j = 1}^{\infty}\sum_{n = j}^{\infty} x^n =$ $\dfrac{a_{1}}{1 - x} + \dfrac{a_{2}}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{a_{1}}{1 - x} + \dfrac{a_{2}}{1 - x}(\dfrac{x}{1 - x}) = \boxed{\dfrac{a_{1}}{1 - x} + \dfrac{a_{2}x}{(1 - x)^2}}$

and,

$\sum_{n = 1}^{\infty} n^2 x^n = \sum_{j = 1}^{\infty}\sum_{n = j}^{\infty} n x^n =$ $\sum_{j = 1}^{\infty} (j x^{j} + (j + 1)x^{j + 1} + (j + 2)x^{j + 2} + ... ) =$ $\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j x^{j} + \dfrac{x}{(1 - x)^2}\sum_{j = 1}^{\infty} x^{j} =$ $(\dfrac{1}{1 - x})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x}{(1 - x)^2})(\dfrac{x}{1 - x})) = \boxed{\dfrac{x^2 + x}{(1 - x)^3}}$

$\implies \boxed{g(x) = \dfrac{a_{1}}{1 - x} + \dfrac{a_{2}x}{(1 - x)^2} + \dfrac{a_{3}(x^2 + x)}{(1 - x)^3}}$

Let $f(x) = \dfrac{4}{9}x^2 + \dfrac{16}{9}x + 1$ .

$f(\dfrac{1}{2}) = g(\dfrac{1}{2})$ and $f(-\dfrac{1}{2}) = g(-\dfrac{1}{2}) \implies$

$\boxed{a_{1} + a_{2} + 3a_{3} = 1}$

$\boxed{9a_{1} - 3a_{2} - a_{3} = 3}$

$\displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) dx = \dfrac{4x^3}{27} + \dfrac{16x^2}{18} + x|_{-\frac{1}{2}}^{\frac{1}{2}} = \dfrac{28}{27}$

$\implies \dfrac{28}{27} = a_{1}\displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{1}{1 - x} dx + a_{2}\displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{x}{(1 - x)^2} dx + a_{3}\displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{x(x + 1)}{(1 - x)^3} dx$

Let $I_{1} = \displaystyle\int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{1}{1 - x} dx = -\ln(1 - x)|_{-\frac{1}{2}}^{\frac{1}{2}} = \boxed{\ln(3)}$

Let $I_{2}(x) = \displaystyle\int \dfrac{x}{(1 - x)^2} dx$ and $u = 1 - x \implies dx = -du \implies$

$I_{2}(x) = -\displaystyle\int u^{-2} - \dfrac{1}{u} du = \dfrac{1}{u} + \ln(u) = \dfrac{1}{1 - x} + \ln(1 - x)$

and,

$I_{2} = I_{2}(x)|_{-\frac{1}{2}}^{\frac{1}{2}} = \boxed{\dfrac{4}{3} - \ln(3)}$

Let $I_{3}(x) = \displaystyle\int \dfrac{x(x + 1)}{(1 - x)^3} dx$ and $u = 1 - x \implies dx = -du \implies$

$I_{3}(x) = - \displaystyle\int \dfrac{1}{u} - 3u^{-2} + 2u^{-3} du = -(\ln(u) + \dfrac{3}{u} - \dfrac{1}{u^2}) = -(\ln(1 - x) + \dfrac{3}{1 - x} - \dfrac{1}{(1 - x)^2})$

and,

$I_{3} = I_{3}(x)|_{-\frac{1}{2}}^{\frac{1}{2}} = \boxed{\ln(3) - \dfrac{4}{9}}$

$\implies$

$$

$$

$R_{1}: \boxed{a_{1} + a_{2} + 3a_{3} = 1}$

$R_{2}: \boxed{9a_{1} - 3a_{2} - a_{3} = 3}$

$R_{3}: \boxed{27\ln(3)a_{1} + 9(4 - 3\ln(3))a_{2} + 3(9\ln(3) - 4)a_{3} = 28}$

$3 * R_{1} + R_{2}$

$(12 - 9\ln(3)) * R_{2} + R_{3}$

$\implies$

$$

$R^{*}_{1}: 6a_{1} + 4a_{3} = 3$

$R^{*}_{2}: (108 - 54\ln(3))a_{1} + (36\ln(3) - 24)a_{3} = 64 - 27\ln(3)$

$(9 - 18\ln(3)) * R^{*}_{1} + R^{*}_{2} \implies \boxed{a_{3} = \dfrac{5}{12(3\ln(3) - 4)}}$

and,

$(6 - 9\ln(3)) * R^{*}_{1} + R^{*}_{2} \implies \boxed{a_{1} = \dfrac{27\ln(3) - 41}{18(3\ln(3) - 4)}}$

and substituting $a_{1}$ and $a_{3}$ into any of the three initial equations $\implies$ $\boxed{a_{2} = \dfrac{54\ln(3) - 107}{18(3\ln(3) - 4)}}$

$\implies a_{1} + a_{2} + a_{3} = \dfrac{18\ln(3) - 29}{6(3\ln(3) - 4)} = \dfrac{2 * 3^2\ln(3) - 29}{2 * 3(3\ln(3) - 2^2)} =$ $\dfrac{\alpha * \beta^2\ln(\beta) - \lambda}{\alpha * \beta(\beta\ln(\beta) - \alpha^{\alpha})}$

$\implies \alpha + \beta + \lambda = \boxed{34}$ .